f'(x)=3ax²+2bx+c
so
12a-4b+c=0
4a/3+4b/3+c=0
解得
b=2a,c=-4a
f'(x)=3ax²+4ax+-4a=a(x+2)(2x-3)
f(x)=ax^3+2ax²-4ax
f(-2)=-8a+8a+8a=8a=-8
a=-1
f(x)=-x^3-2x²+4x
X∈[-3 ,3]
f(-3)=27-18-12=-3
f(3)=-27-18+12=-33
f(2/3)=40/27
在[-3,3],f(x)min=-33,f(x)max=40/27
对X∈[-3 ,3]都有f(x)≥m∧2-14m恒成立
m²-14m≤-33
m²-14m+33≤0
3≤m≤11